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Lemur [1.5K]
3 years ago
13

In order to make an electromagnet, you need a

Physics
2 answers:
Orlov [11]3 years ago
8 0
A battery, some wire, and a nail.
Licemer1 [7]3 years ago
7 0
All you need to do is wrap some insulated copper<span> wire around an </span>iron<span> core. If you attach a battery to the wire, an electric current will begin to flow and the iron core will become magnetized. When the battery is disconnected, the iron core will lose its magnetism.</span>
You might be interested in
What is the wavelength of a radar signal that has a frequency of 27 GHz? The speed of light is 3 × 108 m/s. Answer in units of m
marusya05 [52]

Explanation:

speed of light= c

wave length= L

frequency= f

c=Lf → L= c/f → L= 3 × 10⁸/ 27 × 10⁹ → L = 1/90 ≈ 0.011 m

4 0
2 years ago
Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all
Blababa [14]

Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

6 0
4 years ago
Bernoulli's principle is responsible for most of the lift produced by an airplane wing.
Rina8888 [55]

Answer:

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5 0
3 years ago
Which is an si metric unit of measurement that is used to record the heat transfer of a solution in a classroom investigation?
kumpel [21]
The SI unit for heat energy is joule
4 0
3 years ago
A 150-W lamp is placed into a 120-V ac outlet. What is the peak current?
devlian [24]

Answer:

Explanation:

Formula

W = I * E

Givens

W = 150

E = 120

I = ?

Solution

150 = I * 120   Divide by 120

150/120 = I

5/4  = I

I = 1.25

Note: This is an edited note. You have to assume that 120 is the RMS voltage in order to go any further. That means that the peak voltage is √2 times the size of 120. The current has the same note applied to it. If the voltage is its rms value, then the current must (assuming the properties of the bulb do not change)

On the other hand, if the voltage is the peak value at 120 then 1.25 will be correct.

However I would go with the other answerer's post and multiply both values by  √2

3 0
4 years ago
Read 2 more answers
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